(1-2x)^12 find the first three terms of this expansion?

Question

i dont understand this question, so i dont know what i am meant to do or how i am emant to do it, please could someone help me?

i)Find the first three tersm of the expansion, in acsending poers of x, of (1-2x)^12

ii)hence find the coefficient of x^2 in the expansion of (1+3x)(1-2x)^12

thsnk you guys

Answer 1

i)
Let y = 2x
(1 - y)^12
= 1 - 12y + ((12).(11) / 2!).y²
= 1 - 12.y + 66.y² + -------
(1 - 2x)^12 = 1 - 24x + 264.x² --------

ii)
= (1 + 3x).(1 - 24x + 264.x² - 72.x² -------------
= 1 - 24x + 264x² - 72x² ---------
= 1 - 24x + 192 x²-----------
Coefficient of x² = 192

Answer 2

Hi,

On an expansion of the form (a - b)^12, your first 3 terms would be:

a^12 - 12a^11b + 66a^10b^2

That's because the first term in the expansion is always the first term raised to the 12th power, followed by terms where the first term's exponent decreases by one each term while the exponent of the second term increases each term by 1. So a goes from a^12 to a^11 to a^10 while b goes from b^0 to b^1 to b^2 in the first 3 terms. The first coefficient is 1. Each following coefficient is the previous coefficient times the first exponent in that term divided by the number of the previous term. So the second term's coefficient is the previous coefficient, 1, times the first exponent, 12, divided by that term's number, 1 (first term). Since 1 8 12/1 = 12. the 2nd term has 12 as a coefficient. For the 3rd term, take 2nd term's coefficient times a's exponent divided by term number, or 12 * 11/2 = 66. Signs in front of terms always alternate when there is a negative sign in the problem.

For your problem if you let a = 1 and b = 2x, then
a^12 - 12a^11b + 66a^10b^2 becomes:

(1)^12 - 12(1)^11*2x + 66(1)^10*(2x)^2 =

1 - 24x + 264x^2

These are your first 3 terms.

ii) If you multiply (1 + 3x) (1- 24x + 264x^2), you get:

1 - 24x + 264x^2 + 3x - 72x^2 + 792x^3 which simplifies to:

1 - 21x + 192x^2 + 792x^3

The x^2 term has a coefficient of 192.

I hope that helps!! :-)

Answer 3

You need to know the binomial expansion that (1 + x)^n = 1 + nx + n(n - 1)x²/2 + ......

n = 12 and x is replaced with (-2x)

So (1 - 2x)^12

= 1 + 12(-2x) + 12.11(-2x)²/2 + ...

= 1 - 24x +264x² + ...

(1 + 3x)((1 - 2x)^12

= (1 + 3x)(1 - 24x + 264x² + ...)

= 1 - 24x + 264x² + 3x - 72x² + 792x³ + ....

= 1 - 21x + 192x² + ...

coefficient of x² is 192

Answer 4

(1-2x)^12=1-24x+12.11/2 4x^2+.......
answer=1-24x+264x^2+......
coefficient ofx^2=+264.
coefficient of x^2 in the expansion of (1+3x)(1-2x)^12
=(1-3x)1-24x+264x^2+......
=264+72=336. answer

Answer 5

http://en.wikipedia.org/wiki/Binomial_theorem